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BT
7
4 tháng 6 2018
A = \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)
A = ( 1 - 1/3 ) + ( 1 - 1/15 ) + ( 1 - 1/35 ) + ( 1 - 1/63 ) + ( 1 - 1/99 )
A = ( 1 + 1 + 1 + 1 + 1 ) - ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )
A = 5 - \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)
A = 5 - ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 )
A = 5 - ( 1 - 1/11 )
A = 5 - 10/11
A = 45/11
AK
4 tháng 6 2018
Dấu \(.\)là dấu nhân
\(A=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)
\(\Rightarrow A=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+\left(1-\frac{1}{99}\right)\)
\(\Rightarrow A=\left(1+1+1+1+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(\Rightarrow A=5-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)
\(\Rightarrow A=5-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(\Rightarrow A=5-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(\Rightarrow A=5-\frac{1}{2}.\left(1-\frac{1}{11}\right)\)
\(\Rightarrow A=5-\frac{1}{2}.\frac{10}{11}\)
\(\Rightarrow A=5-\frac{5}{11}\)
\(\Rightarrow A=\frac{55}{11}-\frac{5}{11}\)
\(\Rightarrow A=\frac{50}{11}\)
~ Ủng hộ nhé
ND
1
11 tháng 5 2019
Dấu chấm là dấu nhân
\(P=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)
\(P=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+\left(1-\frac{1}{99}\right)\)
\(P=\left(1+1+1+1+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(P=5-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)
\(P=5-\frac{1}{2}.2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)
\(P=5-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(P=5-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(P=5-\frac{1}{2}.\left(1-\frac{1}{11}\right)\)
\(P=5-\frac{1}{2}.\frac{10}{11}\)
\(P=5-\frac{5}{11}\)
\(P=\frac{55}{11}-\frac{5}{11}\)
\(P=\frac{50}{11}\)
NH
30 tháng 3 2019
\((\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99})x=\frac{2}{3}\)
Đặt \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{9.11}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(A=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)
Thay A vào biểu thức
\(\Rightarrow\frac{5}{11}x=\frac{2}{3}\)
\(\Rightarrow x=\frac{22}{15}\)
P/s: Có thể tính sai :(
30 tháng 3 2019
\(\left[\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right]\times x=\frac{2}{3}\)
Trước tiên mình tính dãy có dấu ngoặc đã
Đặt : \(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(=\frac{1}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right]\)
\(=\frac{1}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right]\)
\(=\frac{1}{2}\left[1-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{11}\right]\)
\(=\frac{1}{2}\left[1-\frac{1}{11}\right]=\frac{1}{2}\cdot\frac{10}{11}=\frac{1\cdot10}{2\cdot11}=\frac{1\cdot5}{1\cdot11}=\frac{5}{11}\)
Thay vào biểu thức \(S=\frac{5}{11}\)ta lại có :
\(\frac{5}{11}\times x=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}:\frac{5}{11}\)
\(\Leftrightarrow x=\frac{2}{3}\cdot\frac{11}{5}\)
\(\Leftrightarrow x=\frac{22}{15}\)
Vậy \(x=\frac{22}{15}\)
17 tháng 2 2015
A= 2( 1/15 + 1/35 + 1/63+ 1/99+1/143)
A= 2(1/3x5 +1/5x7 + 1/7x9 + 1/9x11 + 1/11x13)
A=2(1/3-1/5+1/5-1/+1/7-1/9+1/9-1/11+1/11-1/13)
A=2(1/3-1/13)
A=2x10/39
A=20/39
2 tháng 9 2017
A = 2/15 + 2/35 + 2/63 + 2/99 + 2/143
A = 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + 2/11x13
A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13
A = 1/3 - 1/13
A = 12/13
NT
3
30 tháng 6 2019
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\)
HL
30 tháng 6 2019
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)( chỗ này tối giản nha )
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
~ Hok tốt ~
TQ
2
DD
1
+ \(\frac{1}{n\times\left(n+2\right)}=\frac{\left(n+2\right)-n}{n\times\left(n+2\right)}\)
\(=\frac{n+2}{n\times\left(n+2\right)}-\frac{n}{n\times\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+2}\)
+ \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)
\(=1-\frac{1}{3}+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+1-\frac{1}{99}\)
\(=5-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(=5-\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\)
\(=5-\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(=5-\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)
\(=5-\frac{1}{2}+\frac{1}{22}=\frac{50}{11}\)
=50/11